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yKpoI14uk [10]
3 years ago
5

HELP ASAP!!! 2C4H10(g) + 1302 ===> 8CO2(g) + 10H2O(g)

Chemistry
1 answer:
Ivan3 years ago
6 0

Answer:

30.34g (corrected to 4 significant figures).

Explanation:

Take the atomic mass of C=12.0, H=1.0, O=16.0.

no. of moles = mass / molar mass

So, no. of moles of butane reacted = 10 / (12x4 + 1x10)

= 0.172414 mol

Since O2 is in excess and butane is the limiting reagent, the no. of moles of carbon dioxide produced depends on the no. of moles of butane reacted.

From the equation, the mole ratio of butane:Carbon dioxide = 2: 8 = 1: 4,

meaning 1 mole of butane gives 4 moles of CO2.

Using this ratio,

we can deduce that the no. of moles of CO2 produced =  0.172414 x 4

=0.689655 mol

As mass = no. of moles x molar mass

mass of CO2 produced =   0.689655 x (12.0+16.0x2)

=30.34g (corrected to 4 significant figures).

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The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane wi
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Answer:

See explanation

Explanation:

Number of moles of each gas is

Nitrogen = 15/28 = 0.536 kmoles

Helium = 5/4  = 1.25 kmoles

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Ethane =    20/30 = 0.67 kmoles

Total number of moles =  0.536 kmoles + 1.25 kmoles + 3.75 kmoles +  0.67 kmoles = 6.206 kmoles

Mole fraction of each gas;

Nitrogen = 0.536 kmoles/6.206 kmoles = 0.086

Helium = 1.25 kmoles/6.206 kmoles = 0.201

Methane = 3.75 kmoles/6.206 kmoles =0.604

Ethane = 0.67 kmoles/6.206 kmoles =0.108

Partial pressure of each gas;

Nitrogen = 0.086 * 1200 kPa = 103.2 kPa

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Ethane = 0.108 * 1200 kPa = 129.6 kPa

Apparent specific heat at constant pressure;

Cp = (0.15 * 1.039) + (0.05 * 5.1926) + ( 0.6 * 2.2537) + (0.2 * 1.7662)

Cp = 2.12 KJ Kg-1 K-1

Cv = Cp- Ru/M

Cv= 2.12 - 8.314/16.12 = 1.604 KJ Kg-1 K-1

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